Does there exist a function for every antiderivative?I'm interpreting your question to mean "does every function have an antiderivative?", which isn't quite what you wrote, but I think it's what you meant.
For continuous functions, the answer is yes. If you start with any continuous function f(x) and want to find an antiderivative for it, you can look at the definite integral
x
F(x) = Integral f(t) dt.
0
One form of the fundamental theorem of calculus says that derivative of this is f(x). (F(x) is the area under under the graph of f and above the interval from 0 to x. If you ask what is the rate of change of this area as x increases, the answer is exactly the height of the graph at x, that is, f(x)).
So, F(x) is an antiderivative of f(x). And, the theory of definite integrals guarantees that F(x) exists and is differentiable, as long as f is continuous.
Most functions you normally encounter are either continuous, or else continuous everywhere except at a finite collection of points. For any such function, an antiderivative always exists except possibly at the points of discontinuity.
For more exotic functions without these kinds of continuity properties, it is often very difficult to tell whether or not an antiderivative exists. But such functions don't normally arise in practice.
How about when you are given a derivative of an unknown function and asked to intergrate it. Is there always an answer no matter how complex? Or are there some equations which may never serve as derivatives?There is always an answer (there is always a function whose derivative is the function given to you, provided it is continuous).
However, it may not be possible to express the answer in terms of familiar functions and operations. For example, the antiderivative of e^(x^2) exists, but there is no simpler way to write the function other than to simply say "the antiderivative of e^(x^2)". You can't find a formula for it in terms of familiar functions, but it exists nonetheless.
Is there any better way of expressing the antiderivative of e^(x^2)?No; it cannot be expressed any more simply than that.
In some situations (especially in probability) one needs to work with the antiderivative of e^(-x^2) a lot. Actually, in practical examples what comes up a lot is the antiderivative of 2 / (sqrt pi) e^(x^2). Because it occurs so frequently, it has been given a name, called the "error function", denoted erf(x).
Using complex numbers, you can express the antiderivative of e^(x^2) in terms of this "erf" function: it is (sqrt pi)/(2i) erf(ix) .
But all this is doing is relating the antiderivative of e^(x^2) to another antiderivative that also cannot be expressed any more simply, but which comes up frequently enough that people have given a special name to it.